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In Theorem 27.1 in Topology by Munkres, he states "Let $X$ be a simply ordered set having the least upper bound property. In the order topology, each closed interval in $X$ is compact."

(The LUB property is if a subset is bounded above, then it has a LUB.)

I don't understand how you could have a simply ordered set (a chain) WITHOUT the LUB property. If a subset is bounded and it is a chain, then how can it not have a LUB? Can someone give an example?

Thanks!

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About the quote: $\mathbb{R}$ has the lub and the usual topology induced by the absolute value coincides with the order topology. How is $[0,+\infty)$ compact? –  1015 Apr 11 '13 at 22:37
    
@julien: I guess closed interval means an interval of the form $[a,b]$ where $a$ and $b$ are elements of the poset. –  Damian Sobota Apr 11 '13 at 22:41
    
@julien: Munkres distinguishes between intervals and rays. –  Cameron Buie Apr 11 '13 at 22:41
    
Thank you Cameron and Damian. It would never occur to me not to call a ray an interval. But now that I think about the word "inter-val" it makes sense: $\infty$ is not a val-ue of the set. I think I'll keep calling rays intervals, though. –  1015 Apr 11 '13 at 22:44
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7 Answers

$\Bbb Q$

Added: $\{q\in\Bbb Q:q<\sqrt2\}$ is a chain, it’s bounded above by $2$, say, and it has no least upper bound in $\Bbb Q$, because $\sqrt2$ is irrational.

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@Kara: I like the short answer, but I’ll be happy to expand it if you need me to. –  Brian M. Scott Apr 11 '13 at 22:24
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@Asaf: I’ll give the safe answer: yes! –  Brian M. Scott Apr 11 '13 at 22:38
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@AsafKaragila That's because I only have $6$ accounts. –  1015 Apr 11 '13 at 22:38
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Perhaps this is more appropriate to Meta. Is there a least upper bound on the number of accounts we can have? –  Jay Apr 11 '13 at 22:42
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I think this post requires a reference to one of the $1327$ MSE proofs that $\sqrt{2}$ is irrational. –  1015 Apr 11 '13 at 22:51
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Consider $(-1,0)\cup(0,1)$ with the natural ordering (which is linear). $(-1,0)$ has no least upper bound although it is bounded (eg. by $1/2$).

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That's a very down-to-earth example, +1. –  1015 Apr 11 '13 at 22:48
    
Actually, according to the Wikipedia link given by the OP, the definition it gives for real numbers allows the least upper bound to be included or disincluded from the set it bounds, so the $(-1,0)$ does strictly speaking have a LUB by the OP's given definition. –  AJMansfield Apr 12 '13 at 2:21
    
That's only the real number version, though. –  AJMansfield Apr 12 '13 at 2:37
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@AJMansfield: $(-1,0)$ has a lub as a subset of reals, but not as a subset of $(-1,0)\cup(0,1)$. –  Damian Sobota Apr 12 '13 at 8:46
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$\{0,1\}\times\Bbb Z$ ordered lexicographically. The subset $\{0\}\times\Bbb N$ is bounded from above but has no least upper bound.

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Up to isomorphism, the minimal example is the set $$ \left\{-\frac{1}{n}\;;\;n\geq 1\right\}=-\frac{1}{\mathbb{N}^*}\quad\mbox{in}\quad\frac{1}{\mathbb{Z}^*}=\left\{\frac{1}{n}\;;\;|n|\geq 1\right\} $$ with the obvious ordering.

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@AsafKaragila Really? My ignorance in set theory will kill me... By $S^*$ I mean $S\setminus\{0\}$ whenever there is a $0$. –  1015 Apr 11 '13 at 23:09
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While some do consider me as the king of all existence, I have yet to pass a law requiring everyone to know set theory in a high level, so there is no danger for your life... yet! –  Asaf Karagila Apr 11 '13 at 23:17
    
@AsafKaragila Now that I know the king of all existence, I can die anyway. –  1015 Apr 11 '13 at 23:36
    
It is implicit in your comment that you acknowledge this title of mine. That's nice. Now there's finally one person who acknowledges that! –  Asaf Karagila Apr 11 '13 at 23:38
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According to the strict definitions given by the OP, the null set fails to have a Least Upper Bound while still being simply ordered.

The Least Upper Bound of a set, as defined at the Wikipedia page he links to requires that it be a member of that set. The null set, having no members, clearly lacks a LUB.

However, the definition given for being simply ordered does not require that the set have any elements. Indeed, a set can only lack the property if it has a pair of elements that are not comparable.

So, the null set is indeed Simply Ordered without having the Least Upper Bound property.

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Perhaps your source of confusion is this: if a chain has a greatest element then that serves as a least upper bound, since it is an upper bound (since it is greatest) and is less than or equal to any upper bound for the chain because it's an element of it.

However! Not every chain need have a greatest element! Here's how you might think about this:

Depending on your level of pedantry, you may regard the empty set as a chain (every pair of elements from it is orderable, after all). Then certainly it doesn't have a greatest element, since it has no elements at all. Does it have a least upper bound? Well, every element of the poset is trivially an upper bound for the empty set, so a least upper bound is precisely a least element. We'll assume the poset itself is non-empty, so that the empty set does have an upper bound. Let's call it $u_0$. How can this upper bound fail to be least? Well, there must be another upper bound smaller than it, say $u_1 < u_0$. Likewise, there must be another upper bound smaller than that, and so on: you must have an infinite descending chain $u_0 > u_1 > u_2 > \dots$ and as long as there's nothing below the entire chain, there can't be a least element. The negative integers are the most obvious example.

So if you accept the empty chain as a chain, in the simply ordered set of the negative integers it has upper bounds but no least upper bound.

But that's not terribly satisfying. What about non-empty chains? Well, suppose your poset has an element $x_0$. One of two things can happen: either $x_0$ is maximal for the chain, or some $x_1$ has $x_1 > x_0$. Is $x_1$ maximal? Maybe, or maybe there's an $x_2 > x_1$, and so on. If this set is to avoid having a maximal element (hence a least upper bound) it must go upwards forever – an ascending chain. It should be bounded – okay, just invent an element that goes above the entire ascending chain. But it can't have a least upper bound – this precisely means the set of upper bounds can't have a least element, and as we discussed before that basically means we need an infinite descending chain.

So if we have an infinite ascending chain of elements, and an infinite descending chain of upper bounds for it, we'll fail to have a least upper bound again.

The lesson to be learnt here is that posets are very flexible structures, and you can form one by just listing elements and placing them relative to each other: it's not like, say, a group, where there is quite a lot of rigidity to the structure, or a field, where you get barely any choice at all.

Indeed, one possible approach would be to take the real numbers, find a subset $C$ with a least upper bound $u$, but $u \not \in C$. Form a new poset by removing $u$ from the reals. Now $C$ doesn't have a least upper bound anymore!

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I just realized that my answer is the same argument as the first half of your third paragraph. I love these degenerate solutions. –  AJMansfield Apr 12 '13 at 2:40
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One more answer, not discussed till now.

Consider $A = \{x,y,z\}$. So $\mathbb{P}(A)= \{\phi, \{x\}, \{y\}, \{z\}, \{x, y\}, \{y,z\}, \{x, z\}, A\}$.

Consider partial order relation $<$ on $\mathbb{P}(A)$ as $a<b$ iff $a\subset b$.

Consider the element $\phi \in \mathbb{P}(A)$. See nay element of $\mathbb{P}(A)$ is an upper bound of the set $\phi$. What is the least upper bound ? Not exists.

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What is wrong here? Please give reason for this downvote. –  Dutta Feb 18 at 4:18
    
It's not a simply/linear/total order the relation you propose, so the answer must be changed, because it's not what the OP asked. Even though I don't agree on givin down votes without explanation :/ –  Riccardo Mar 1 at 15:09
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